Chapter 3 Methods of Analysis: 2) Mesh Analysis Dr. Waleed Al-Hanafy waleed alhanafy@yahoo.com Faculty of Electronic Engineering, Menoufia Univ., Egypt MSA Summer Course: Electric Circuit Analysis I (ESE 233) Lecture no. 5 July 27, 2011
Overview 1 Mesh Analysis Procedures 2 Mesh Analysis with Current Sources 3 Conclusions Reference: [1] Alexander Sadiku, Fundamentals of Electric Circuits, 4th ed. McGraw-Hill, 2009.
Introduction Mesh analysis provides another general procedure for analysing circuits, using mesh currents as the circuit variables Using mesh currents instead of element currents as circuit variables is convenient and reduces the number of equations that must be solved simultaneously Recall that a loop is a closed path with no node passed more than once. A mesh is a loop that does not contain any other loop within it Nodal analysis applies KCL to find unknown voltages in a given circuit, while mesh analysis applies KVL to find unknown currents Mesh analysis is not quite as general as nodal analysis because it is only applicable to a circuit that is planar A planar circuit is one that can be drawn in a plane with no branches crossing one another; otherwise it is nonplanar A circuit may have crossing branches and still be planar if it can be redrawn such that it has no crossing branches
Example
What is a Mesh? A mesh is a loop which does not contain any other loops within it Paths abefa and bcdeb are meshes, but path abcdefa is not a mesh
Mesh Method Steps Steps to Determine the Mesh Currents: 1 Assign mesh currents i 1, i 2,, i n to the n meshes 2 Apply KVL to each of the n meshes. Use Ohm s law to express the voltages in terms of the mesh currents 3 Solve the resulting n simultaneous equations to get the mesh currents.
Example-1 Find the branch currents I1, I2, and I3 using mesh analysis For mesh 1, applying KVL 15 + 5i 1 + 10(i 1 i 2 ) + 10 = 0 For mesh 2, or 3i 1 2i 2 = 1 6i 2 + 4i 2 + 10(i 2 i 1 ) 10 = 0 or i 1 = 2i 2 1 Solving for i 1 and i 2 results in i 1 = i 2 = 1 A. Thus I 1 = i 1 = 1 A, I 2 = i 2 = 1 A, and I 3 = i 1 i 2 = 0
Exercise-1 Calculate the mesh currents i 1 and i 2 in the circuit shown Answer: i 1 = 2 3 A, i 2 = 0
Example-2 Use mesh analysis to find the current i 0 in the circuit shown For mesh 1, 24 + 10(i 1 i 2 ) + 12(i 1 i 3 ) = 0 = 11i 1 5i 2 6i 3 = 12 For mesh 2, 24i 2 + 4(i 2 i 3 ) + 10(i 2 i 1 ) = 0 = 5i 1 + 19i 2 2i 3 = 0 For mesh 3, 4i 0 + 12(i 3 i 1 ) + 4(i 3 i 2 ) = 0 Since i 0 = i 1 i 2, then = i 1 i 2 + 2i 3 = 0 Finally: i 1 = 2.25 A, i 2 =.75 A, and i 3 = 1.5 A, thus i 0 = i 1 i 2 = 1.5 A
Applying mesh analysis to circuits containing current sources (dependent or independent) may appear complicated. But it is actually much easier than what we encountered in the previous section, because the presence of the current sources reduces the number of equations. Case 1: When a current source exists only in one mesh: Consider the circuit below, we set i 2 = 5 A and write a mesh equation for the other mesh in the usual way, that is, 10 + 4i 1 + 6(i 1 + i 2 ) = 0 i 1 = 2 A
Case 2: When a current source exists between two meshes: Consider the circuit below we create a supermesh by excluding the current source and any elements connected in series with it, as shown in Fig. (b). Thus, A supermesh results when two meshes have a (dependent or independent) current source in common. Thus 20 + 6i 1 + 10i 2 + 4i 2 = 0 6i 1 + 14i 2 = 20. Since by applying KCL at node 0, i 2 = i 1 + 6, then i 1 = 3.2 A and i 2 = 2.8 A.
Example-3 For the circuit shown, find i 1 to i 4 using mesh analysis Note that meshes 1 and 2 form a supermesh since they have an independent current source in common. Also, meshes 2 and 3 form another supermesh because they have a dependent current source in common. The two supermeshes intersect and form a larger supermesh as shown. For the larger supermesh, we have 2i 1 + 4i 3 + 8(i 3 i 4) + 6i 2 = 0. We also have at node P, i 2 = i 1 + 5 and at node Q i 2 = i 3 + 3i 0. But i 0 = i 4. At mesh 4, 2i 4 + 8(i 4 i 3) + 10 = 0. Solving results in i 1 = 7.5 A, i 2 = 2.5 A, i 3 = 3.93 A, and i 2 = 2.143 A.
Homework Use mesh analysis to determine i 1, i 2, and i 3 in the circuit shown Answer: i 1 = 3.474 A, i 2 =.4737 A, and i 3 = 1.1052 A
Conclusion Concluding remarks Mesh analysis method is studied as a key tool to analyse any circuit Basic mesh analysis steps is introduced highlighted by some examples The case of supermesh is also given with examples.